NEET Test Series from KOTA - 10 Papers In MS WORD
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CHXI07:EQUILIBRIUM
314719
The degree of hydrolysis and \(\mathrm{pH}\) of salts made up of weak acid and weak base is independent of
1 \({\rm{p}}{{\rm{K}}_{\rm{a}}}\)
2 \({\rm{p}}{{\rm{K}}_{\rm{b}}}\)
3 concentration of salt
4 \({\rm{p}}{{\rm{K}}_{\rm{w}}}\)
Explanation:
\(\mathrm {p H=7+1 / 2\left[p K_{a}+p K_{b}\right]}\) which is independent of concentration of salt
CHXI07:EQUILIBRIUM
314720
A weak acid \({\text{HX}}\) has the dissociation constant \(1 \times 10^{-5} \mathrm{M}\). It forms a salt \(\mathrm{NaX}\) on reaction with alkali. The degree of hydrolysis of \(0.1 \mathrm{M}\) solution of \(\mathrm{NaX}\) is
1 \(0.0001 \%\)
2 \(0.01 \%\)
3 \(0.1 \%\)
4 \(0.15 \%\)
Explanation:
\(\mathrm{X}^{-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{HX}+\mathrm{OH}^{-}\) \(\mathrm{K}_{\mathrm{h}}=\dfrac{\mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{a}}}=\dfrac{10^{-14}}{10^{-5}}=10^{-9}\) So, degree of hydrolysis \(\mathrm{h}=\sqrt{\dfrac{\mathrm{K}_{\mathrm{h}}}{\mathrm{C}}}=\sqrt{\dfrac{10^{-9}}{10^{-1}}}=10^{-4}\) Percentage of degree of hydrolysis \(=10^{-4} \times 100=10^{-2} \%=0.01 \%\)
AIIMS - 2012
CHXI07:EQUILIBRIUM
314721
Calculate the hydrolysis constant of the salt containing \(\mathrm{NO}_{2}^{-}\). Given \(\mathrm{K}_{a}\) for \(\mathrm{HNO}_{2}=4.5 \times 10^{-10}\).
314722
The aqueous solution of \(\mathrm{HCOONa}\), \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+} \mathrm{Cl}^{-}\)and \(\mathrm{KCN}\) are respectively
314719
The degree of hydrolysis and \(\mathrm{pH}\) of salts made up of weak acid and weak base is independent of
1 \({\rm{p}}{{\rm{K}}_{\rm{a}}}\)
2 \({\rm{p}}{{\rm{K}}_{\rm{b}}}\)
3 concentration of salt
4 \({\rm{p}}{{\rm{K}}_{\rm{w}}}\)
Explanation:
\(\mathrm {p H=7+1 / 2\left[p K_{a}+p K_{b}\right]}\) which is independent of concentration of salt
CHXI07:EQUILIBRIUM
314720
A weak acid \({\text{HX}}\) has the dissociation constant \(1 \times 10^{-5} \mathrm{M}\). It forms a salt \(\mathrm{NaX}\) on reaction with alkali. The degree of hydrolysis of \(0.1 \mathrm{M}\) solution of \(\mathrm{NaX}\) is
1 \(0.0001 \%\)
2 \(0.01 \%\)
3 \(0.1 \%\)
4 \(0.15 \%\)
Explanation:
\(\mathrm{X}^{-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{HX}+\mathrm{OH}^{-}\) \(\mathrm{K}_{\mathrm{h}}=\dfrac{\mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{a}}}=\dfrac{10^{-14}}{10^{-5}}=10^{-9}\) So, degree of hydrolysis \(\mathrm{h}=\sqrt{\dfrac{\mathrm{K}_{\mathrm{h}}}{\mathrm{C}}}=\sqrt{\dfrac{10^{-9}}{10^{-1}}}=10^{-4}\) Percentage of degree of hydrolysis \(=10^{-4} \times 100=10^{-2} \%=0.01 \%\)
AIIMS - 2012
CHXI07:EQUILIBRIUM
314721
Calculate the hydrolysis constant of the salt containing \(\mathrm{NO}_{2}^{-}\). Given \(\mathrm{K}_{a}\) for \(\mathrm{HNO}_{2}=4.5 \times 10^{-10}\).
314722
The aqueous solution of \(\mathrm{HCOONa}\), \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+} \mathrm{Cl}^{-}\)and \(\mathrm{KCN}\) are respectively
314719
The degree of hydrolysis and \(\mathrm{pH}\) of salts made up of weak acid and weak base is independent of
1 \({\rm{p}}{{\rm{K}}_{\rm{a}}}\)
2 \({\rm{p}}{{\rm{K}}_{\rm{b}}}\)
3 concentration of salt
4 \({\rm{p}}{{\rm{K}}_{\rm{w}}}\)
Explanation:
\(\mathrm {p H=7+1 / 2\left[p K_{a}+p K_{b}\right]}\) which is independent of concentration of salt
CHXI07:EQUILIBRIUM
314720
A weak acid \({\text{HX}}\) has the dissociation constant \(1 \times 10^{-5} \mathrm{M}\). It forms a salt \(\mathrm{NaX}\) on reaction with alkali. The degree of hydrolysis of \(0.1 \mathrm{M}\) solution of \(\mathrm{NaX}\) is
1 \(0.0001 \%\)
2 \(0.01 \%\)
3 \(0.1 \%\)
4 \(0.15 \%\)
Explanation:
\(\mathrm{X}^{-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{HX}+\mathrm{OH}^{-}\) \(\mathrm{K}_{\mathrm{h}}=\dfrac{\mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{a}}}=\dfrac{10^{-14}}{10^{-5}}=10^{-9}\) So, degree of hydrolysis \(\mathrm{h}=\sqrt{\dfrac{\mathrm{K}_{\mathrm{h}}}{\mathrm{C}}}=\sqrt{\dfrac{10^{-9}}{10^{-1}}}=10^{-4}\) Percentage of degree of hydrolysis \(=10^{-4} \times 100=10^{-2} \%=0.01 \%\)
AIIMS - 2012
CHXI07:EQUILIBRIUM
314721
Calculate the hydrolysis constant of the salt containing \(\mathrm{NO}_{2}^{-}\). Given \(\mathrm{K}_{a}\) for \(\mathrm{HNO}_{2}=4.5 \times 10^{-10}\).
314722
The aqueous solution of \(\mathrm{HCOONa}\), \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+} \mathrm{Cl}^{-}\)and \(\mathrm{KCN}\) are respectively
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
CHXI07:EQUILIBRIUM
314719
The degree of hydrolysis and \(\mathrm{pH}\) of salts made up of weak acid and weak base is independent of
1 \({\rm{p}}{{\rm{K}}_{\rm{a}}}\)
2 \({\rm{p}}{{\rm{K}}_{\rm{b}}}\)
3 concentration of salt
4 \({\rm{p}}{{\rm{K}}_{\rm{w}}}\)
Explanation:
\(\mathrm {p H=7+1 / 2\left[p K_{a}+p K_{b}\right]}\) which is independent of concentration of salt
CHXI07:EQUILIBRIUM
314720
A weak acid \({\text{HX}}\) has the dissociation constant \(1 \times 10^{-5} \mathrm{M}\). It forms a salt \(\mathrm{NaX}\) on reaction with alkali. The degree of hydrolysis of \(0.1 \mathrm{M}\) solution of \(\mathrm{NaX}\) is
1 \(0.0001 \%\)
2 \(0.01 \%\)
3 \(0.1 \%\)
4 \(0.15 \%\)
Explanation:
\(\mathrm{X}^{-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{HX}+\mathrm{OH}^{-}\) \(\mathrm{K}_{\mathrm{h}}=\dfrac{\mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{a}}}=\dfrac{10^{-14}}{10^{-5}}=10^{-9}\) So, degree of hydrolysis \(\mathrm{h}=\sqrt{\dfrac{\mathrm{K}_{\mathrm{h}}}{\mathrm{C}}}=\sqrt{\dfrac{10^{-9}}{10^{-1}}}=10^{-4}\) Percentage of degree of hydrolysis \(=10^{-4} \times 100=10^{-2} \%=0.01 \%\)
AIIMS - 2012
CHXI07:EQUILIBRIUM
314721
Calculate the hydrolysis constant of the salt containing \(\mathrm{NO}_{2}^{-}\). Given \(\mathrm{K}_{a}\) for \(\mathrm{HNO}_{2}=4.5 \times 10^{-10}\).
314722
The aqueous solution of \(\mathrm{HCOONa}\), \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+} \mathrm{Cl}^{-}\)and \(\mathrm{KCN}\) are respectively
